Algebra of Functions, precalculus lesson

A shop’s profit is its revenue minus its costs, and both depend on how many items it sells. Each of these is a function of the same input, and the profit at one sales level comes from subtracting two numbers. Functions can be added, subtracted, multiplied and divided in exactly this way, one input at a time, as long as both functions accept that input.

Key terms

Sum and difference

Definition Let $f$ and $g$ be functions with domains $D_f$ and $D_g$ . Their sum $f + g$ and their difference $f - g$ are the functions given by

\begin{gathered} (f + g)(x) = f(x) + g(x), \\[0.4em] (f - g)(x) = f(x) - g(x), \end{gathered}

for every $x$ in $D_f \cap D_g$ , which is the domain of both. The order matters for the difference, since $(g - f)(x) = -(f - g)(x)$ .

In plain words To add two functions, add their outputs at the same input. The new function makes sense only at inputs where both original functions do.

Example Let $f(x) = x^2 + 1$ and $g(x) = 3x - 5$ , both defined for all real numbers. Then

\begin{aligned} (f + g)(x) &= x^2 + 3x - 4, \\[0.4em] (f - g)(x) &= x^2 + 1 - (3x - 5) \\[0.4em] &= x^2 - 3x + 6. \end{aligned}

At $x = 2$ the outputs are $f(2) = 5$ and $g(2) = 1$ , so $(f + g)(2) = 6$ and $(f - g)(2) = 4$ , as the formulas give. The parentheses around $3x - 5$ matter, because the minus sign applies to the whole output of $g$ .

Product

Definition The product $fg$ of functions $f$ and $g$ is the function given by

(fg)(x) = f(x) \cdot g(x)

for every $x$ in $D_f \cap D_g$ . A constant multiple $cf$ , with $(cf)(x) = c \cdot f(x)$ , is the product of $f$ with a constant function and has domain $D_f$ . The notation $fg$ means multiplying outputs. It is not the composition $f \circ g$ , which feeds the output of $g$ into $f$ .

In plain words Multiply the two outputs at each input. Where either factor is zero, the product is zero.

Example Let $f(x) = 2x - 1$ and $g(x) = x + 3$ . Then

\begin{aligned} (fg)(x) &= (2x - 1)(x + 3) \\[0.4em] &= 2x^2 + 5x - 3. \end{aligned}

At $x = 1$ the outputs are $f(1) = 1$ and $g(1) = 4$ , so $(fg)(1) = 4$ , and the formula gives $2 + 5 - 3 = 4$ . The zeros of $fg$ are the zeros of the two factors, $\frac{1}{2}$ and $-3$ .

Domain intersection

Definition The domain of $f + g$ , $f - g$ and $fg$ is the intersection

\begin{aligned} &D_f \cap D_g \\[0.4em] &\quad = \{\, x \mid x \in D_f \text{ and } x \in D_g \,\}, \end{aligned}

the set of inputs that belong to both domains. It is read from the original formulas of $f$ and $g$ , before any simplification of the combined formula.

In plain words An input is allowed only when both functions accept it. On a number line, the domain is the part where the two domain sets overlap.

Example Let $f(x) = \sqrt{x + 2}$ and $g(x) = \sqrt{5 - x}$ , with domains $[-2, \infty)$ and $(-\infty, 5]$ . Then

\begin{aligned} D_{f + g} &= [-2, \infty) \cap (-\infty, 5] \\[0.4em] &= [-2, 5]. \end{aligned}

At $x = 7$ the value $f(7) = 3$ exists, but $g(7) = \sqrt{-2}$ is not a real number, so $(f + g)(7)$ is undefined. Simplifying can hide this. With $p(x) = \sqrt{x - 1}$ and $q(x) = \sqrt{x + 1}$ the product is $\sqrt{x^2 - 1}$ for $x \ge 1$ , and that formula alone also works at $x = -2$ . Yet $p(-2)$ is not real, so $-2$ is not in the domain $[1, \infty)$ of $pq$ .

Quotient

Definition The quotient $\frac{f}{g}$ is the function given by

\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}

for every $x$ in $D_f \cap D_g$ with $g(x) \ne 0$ . Its domain is therefore

\{\, x \in D_f \cap D_g \mid g(x) \ne 0 \,\}.

In plain words Divide the outputs at each input. Both functions must be defined there, and the bottom one must not be zero.

Example Let $f(x) = x + 4$ and $g(x) = x^2 - 9$ , both defined for all real numbers. The divisor is zero at $x = 3$ and $x = -3$ , so

\begin{gathered} \left(\frac{f}{g}\right)(x) = \frac{x + 4}{x^2 - 9}, \\[0.8em] x \ne -3, \ x \ne 3. \end{gathered}

At $x = -4$ the quotient is $\frac{0}{7} = 0$ , which is perfectly fine. The reversed quotient $\frac{g}{f}$ has a different domain, every real number except $-4$ .

Nonzero divisor

Definition In $\frac{f}{g}$ the function $g$ is the divisor, and an input $x$ is allowed only when $g(x)$ is defined and $g(x) \ne 0$ . The zeros of the divisor are removed from $D_f \cap D_g$ even when $g$ itself is defined there. A zero of the numerator $f$ causes no exclusion by itself. If it lies in $D_g$ and is not a zero of $g$ , it stays in the domain and is a zero of the quotient. If it is also a zero of $g$ , as $x = 1$ is for $\frac{x^2 - 1}{x - 1}$ , it is excluded like every other zero of the divisor.

In plain words Before you divide, find every input where the bottom function is zero and throw it out. A zero on top does no harm by itself.

Example Let $f(x) = x - 2$ and $g(x) = \sqrt{x}$ . The square root is defined at $0$ , but it equals $0$ there, so

\left(\frac{f}{g}\right)(x) = \frac{x - 2}{\sqrt{x}}, \qquad x \in (0, \infty).

The reversed quotient $\frac{\sqrt{x}}{x - 2}$ has domain $[0, 2) \cup (2, \infty)$ . It keeps $0$ , where its value is $\frac{0}{-2} = 0$ . The same condition returns in calculus, where the quotient law for limits needs a denominator whose limit is not zero.

Graph of pointwise operations

Definition The operations above are pointwise, which means each one is carried out separately at each input. For $x$ in the common domain, the point of the graph of $f + g$ above $x$ is $(x, f(x) + g(x))$ , so its height is the sum of the signed heights of the graphs of $f$ and $g$ above that same $x$ . The other operations work the same way. In particular $(f - g)(x) = 0$ exactly when $f(x) = g(x)$ , so the zeros of $f - g$ are the first coordinates of the points where the graphs of $f$ and $g$ meet.

In plain words Stand at one input, read both heights and combine them. A height below the $x$ -axis counts as negative.

Example Take $f(x) = x^2$ and $g(x) = x + 2$ .

$x$	$f(x)$	$g(x)$	$(f + g)(x)$	$(f - g)(x)$
$-2$	$4$	$0$	$4$	$4$
$-1$	$1$	$1$	$2$	$0$
$0$	$0$	$2$	$2$	$-2$
$1$	$1$	$3$	$4$	$-2$
$2$	$4$	$4$	$8$	$0$

The difference is $x^2 - x - 2 = (x + 1)(x - 2)$ , with zeros $-1$ and $2$ , and the graphs of $f$ and $g$ meet at $(-1, 1)$ and $(2, 4)$ . At $x = -2$ , where $g$ is zero, the graph of $f + g$ passes through the point $(-2, 4)$ of the graph of $f$ .

Restrictions survive cancellation

Definition The domain of a combination is fixed by the original functions before any algebra is done. If the formula of $f + g$ , $f - g$ , $fg$ or $\frac{f}{g}$ simplifies, for example by cancelling a common factor or two opposite terms, the simplified formula agrees with the combination only on that domain. Inputs excluded at the start stay excluded, so the graph of the combination is the graph of the simplified formula with those inputs removed. A single excluded input where the simplified formula is defined shows up as a hole in the graph.

In plain words Simplifying changes how the formula looks, not which inputs are allowed. Write the restriction next to the simplified formula so it is not lost.

Example Let $f(x) = x^2 - 1$ and $g(x) = x - 1$ . For $x \ne 1$

\begin{aligned} \left(\frac{f}{g}\right)(x) &= \frac{(x - 1)(x + 1)}{x - 1} \\[0.8em] &= x + 1. \end{aligned}

The graph is the line $y = x + 1$ with a hole at $(1, 2)$ . Cancelling terms works the same way. With $p(x) = x + \frac{1}{x}$ and $q(x) = \frac{1}{x}$ the difference is $(p - q)(x) = x$ , but only for $x \ne 0$ , since neither $p$ nor $q$ is defined at $0$ .

Check yourself

Let $f(x) = \sqrt{x - 1}$ and $g(x) = x - 3$ . Find the domains of $f + g$ , $fg$ , $\frac{f}{g}$ and $\frac{g}{f}$ .

Show the answer

The domains are $D_f = [1, \infty)$ and $D_g = \mathbb{R}$ , so $f + g$ and $fg$ both have domain $[1, \infty)$ . The quotient $\frac{f}{g}$ also excludes the zero $3$ of $g$ , giving $[1, 3) \cup (3, \infty)$ . The quotient $\frac{g}{f}$ excludes the zero $1$ of $f$ , giving $(1, \infty)$ .

Suppose $f(3) = 4$ and $g(3) = -2$ . Find $(f - g)(3)$ , $(fg)(3)$ and $\left(\frac{f}{g}\right)(3)$ .

Show the answer

Combine the two outputs at the input $3$ . This gives $(f - g)(3) = 4 - (-2) = 6$ , then $(fg)(3) = 4 \cdot (-2) = -8$ , and $\left(\frac{f}{g}\right)(3) = \frac{4}{-2} = -2$ .

Let $f(x) = x^2 + 2x$ and $g(x) = x + 2$ . Simplify $\frac{f}{g}$ and describe its graph.

Show the answer

For $x \ne -2$

\frac{x^2 + 2x}{x + 2} = \frac{x(x + 2)}{x + 2} = x.

The value $-2$ stays excluded because $g(-2) = 0$ . The graph is the line $y = x$ with a hole at $(-2, -2)$ .

Algebra of Functions

Key terms

Sum and difference

Product

Domain intersection

Quotient

Nonzero divisor

Graph of pointwise operations

Restrictions survive cancellation

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus